16.1.3 First-Order Linear Equations

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1. First-Order Linear Differential Equations

In this section, the abbreviation “DE” stands for “differential equation”.

In Section 7.5 Part 3, we encountered the DE dy/dt = ky, or y‘ + by = 0, where b = – k. In this equation, the highest order of
derivative is 1 (the equation is of first order), and each term contains at most one factor y or y‘, but not both, whose power is 1
(the equation is linear). It’s called a first-order linear differential equation. Similarly, the equation x²y – 2x³y‘ – e^x + 3 = 0 is a
first-order linear DE. It can be converted to the form y‘ – (1/2x)y = (3 – e^x)/2x³ by changing the order of the first 2 terms,
moving the last 2 terms to the right-hand side, and dividing the resulting equation by -2x³. This form is called the standard
form of the DE.

Definition 1.1 – First-Order Linear Differential Equations

A DE that can be written in the form: y‘ + p(x)y = q(x), where p(x) and q(x) are given continuous functions, is called a first-order linear differential equation. The form y‘ + p(x) y = q(x) is called the standard form of the DE.

The highest derivative order is 1, thus the adjective “first-order”, and each term contains at most either a factor y or a factor y‘
raised to just the power of 1, thus the adjective “linear” (note: a term such as y‘y is of power 1 + 1 = 2). Remark that the
equation is written in the decreasing order of the derivative (recall that y = y⁽⁰⁾). The coefficient of y‘ is 1 and that of y is p(x).
Observe that while the equation is linear, each function y, y‘, p(x), and q(x) doesn’t have to be linear. For example, y = 2x + 3
is linear while y = x² + 3 and q(x) = sin x

The DE y‘ = f (x), as discussed in Section 16.1.1 Part 2, is a first-order linear one. For this equation, p(x) = 0 and q(x) = f (x).

When The Coefficient Of y‘ Isn’t 1

Suppose the coefficient of y‘ of the DE:

f(x)y‘ + p₁(x)y = q₁(x)

is the function f(x) that’s not the constant function 1. In this case, to get an equation of the form y‘ + p(x)y = q(x) that has 1
as the coefficient of y‘ and that’s equivalent to the given equation, we divide both sides of the given equation by f(x):

A Particular Case

Consider the simple example of the particular case of the above DE, y‘ + by = 0, where p(x) = b and q(x) = 0, and where b is
a constant. We solve it as follows:

y‘ + by = 0

y‘ = – by,

So y = Ce^–bx + K isn’t the general solution of y‘ + by = 0. The reason is that the general solution of a first-order equation
involves only one arbitrary constant, and y = Ce^–bx already has one arbitrary constant, which is C. The arbitrary constant C
originates from the arbitrary constant C₁, which is added to –bx to obtain the general antiderivative of –b. This confirms that
we’ve already added an arbitrary constant.

Now let’s solve the DE by using an alternative approach. The left-hand side of y‘ + by = 0 has 2 terms, one term contains the
derivative y‘ as a factor and the other term contains the function y as a factor. This reminds us of the product rule of
differentiation: (uv)’ = u‘v + uv‘, or (uv)’ = uv‘ + u‘v, where the first term of the right-hand side contains the derivative v‘
as a factor and the other term contains the function v as a factor. So we try to examine if y‘ + by is the derivative of the
product of some function, say u, and y, so that the product is uy, whose derivative is (uy)’ = uy‘ + u‘y. We see that y‘ + by =
1y‘ + by, so it should be that u = 1 and u‘ = b. But b isn’t the derivative of 1 unless b = 0. Thus y‘ + by, when b is non-0, in
itself can’t be the derivative of the product of any function and y. The problem is that the coefficient of y‘ is 1. So we multiply
both sides of y‘ + by = 0 by a function u, obtaining uy‘ + buy = 0. In order for uy‘ + buy to be the derivative uy‘ + u‘y of uy,
bu must be the derivative u‘ of u. We determine u so that this requirement is fulfilled, as follows:

u‘ = bu,

Multiplying y‘ + by = 0 by e^bx works out well. The general solution is the same as obtained previously.

General Case

We don’t need to memorize the formula of the general solution y of the DE y‘ + p(x)y = q(x). In solving such an equation,
we instead can follow the procedure that leads to that formula.

Integrating Factors

In the general solution:

The solution procedure of the first-order linear DE can be summarized as follows.

Solution Procedure Of The First-Order Linear Differential Equation

The first-order linear DE of the standard form y‘ + p(x)y = q(x), where p(x) and q(x) are continuous, is solved by the integrating-factor technique as follows (you don’t have to remember things in brackets that start with “thus”): Make sure that the DE is in the standard form y‘ + p(x) y = q(x), where the coefficient of y’ is 1, because the above technique assumes the standard form. If it’s not in the standard form, convert it to the standard form.

Example 1.1

In Section 7.5 Part 3, we solved the DE dy/dt = ky, or y‘ + by = 0, where b = – k, and found the general solution to be y =
Ce^kt, where C = y(0). Use the integrating-factor technique described in this section to show that the general solution of the DE:

y‘ + by = 0,

where y = y(x) is a function of x and b is a constant, is y = y(0)e^–bx. Note that we’ve dealt with this DE under the heading “A
Particular Case” above.

Note

Here the functions p(x) = b and q(x) = 0 are constant and thus continuous.

Solution
An antiderivative of b is bx. Then:

e^bxy’ + e^bxby = 0     (multiply both sides of the given DE by e^bx),
(e^bxy)’ = 0,

y(x) = y = Ce^–bx,
y(0) = Ce⁰ = C(1) = C,
y = y(0)e^–bx.
EOS

Remark 1.1 – On Example 1.1

Let’s find out what happens if we utilize the general antiderivative bx + c of b, where c is an arbitrary constant, instead of the
simple particular antiderivative bx, where c = 0:

e^bx+cy’ + e^bx+cby = 0     (multiply both sides of the given DE by e^bx+c),
(e^bx+cy)’ = 0,

y(x) = y = Ce^–bx–c = Ce^–ce^–bx,
y(0) = Ce^–ce⁰ = Ce^–c(1) = Ce^–c,
y = y(0)e^–bx.

It’s the same general solution. Thus utilizing any particular antiderivative of p(x) is fine. We utilize the simplest one, where
the constant of integration is 0.

Example 1.2

Solve y‘ + y/x = 2, where x > 0.

Solution

EOS

Example 1.3

Solve the first-order linear DE dy/dx + xy = 2x³.

Solution

Example 1.4

Solve this initial-value problem:

Solution

EOS

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We now present examples of some of the applications of the first-order linear DEs.

Concentrations In (Liquid) Solutions

Example 2.1

A tank contains 200 L of brine in which 2.5 kg of salt is dissolved. Then brine containing 0.1 kg of salt per liter enters the tank at a rate of 15 L/min. While the brine is entering the tank, brine from the tank is also exiting the tank at a rate of 8 L/min. The concentration of the brine in the tank is kept uniform throughout the tank by stirring. Find the amount of salt in the tank after 30

min.

Solution
Let S be the amount of salt in the tank at time t, where S is in kg and t in min. The initial time t = 0 is when the brine starts to
enter and exit the tank. The rate of change of the amount of salt in the tank at time t is:

C = (-17.5)(200^8/7),

After 30 min, the amount of salt in the tank is approximately 33.3 kg.

EOS

Remark 2.1 – On Example 2.1

We’re to determine a quantity. First we find the rate of change or derivative of that quantity. We obtain a DE whose unknown
function is the quantity. Then we solve the DE and get the quantity.

Savings Accounts

Example 2.2

A savings account is opened with a deposit of $10,000. Money is being continuously transferred automatically from another account and deposited into this account in such a way that the accumulated deposited amount increases at the rate of (1,000 +

200

t) dollars per year, where t is the number of years after the account is opened. Interest is being paid also continuously into
the account at the rate of 13% per year. Determine the balance of the account after 5 years.

Notes

1. The quantity 1,000 + 200t is the rate of increase of the accumulated deposited amount (ADA), not of the deposit, at time t.
    For example, for the simpler case of an ADA increasing at the constant rate of $5,000 per year, the deposit is constant at
    $5,000 per year and thus increases at the rate of $0 per year.

2. Since the interest is paid continuously, the accumulated interest (AI) at time t increases at (13%)B = $0.13B per year, where
    B is the balance at time t. In this problem the unit of time is the year. So the rate of increase of the AI (not the rate of
    interest, not the rate of increase of the rate of interest, the rate of interest being fixed at 13% per year) is also $0.13B per
    year.

3. Since the deposit is made continuously (the number of deposits on any time interval approaches infinity), we wonder if the
    ADA is infinite and thus the balance is infinite at any time t > 0 (the initial time t = 0 is when the account is opened). Let’s
    find out. As the rate of increase of the ADA is 1,000 + 200t and the initial amount (at time t = 0) is $10,000, the ADA at
    time t is a(t) = 1,000t + 200(t²/2) + 10,000, or a(t) = 100t² + 1,000t + 10,000. So at any time t where 0 < t < infinity, the
    ADA a(t) is finite. Although the parabola a(t) rises or increases continuously for all increasing t > 0, its height at any t
    where 0 < t < infinity is finite. For any bounded interval [t₁, t₂] where 0 < t₁ < t₂, divide [t₁, t₂] into n equal sub-intervals,
    and consider the parabola as increasing n times on [t₁, t₂] from a(t₁) to a(t₂). For the continuous increase, let n approach
    infinity. As the number of increases approaches infinity, each increase approaches 0 and fast, fast enough for their sum to be
    finite. Thus the height of the parabola at any t where 0 < t < infinity is finite. Although the deposits are made continuously,
    each is small enough that the ADA is finite. As for the interest, we already know that although it’s paid continuously, the
    accumulated interest is finite. Thus at any time t where 0 < t < infinity the balance is finite.

Solution
Let B be the balance at time t. Then the rate of increase of the balance at time t is:

Let u = t and dv = e^–0.13t dt, so that du = dt and v = –e^–0.13t/0.13. Then:

The balance of the account after 5 years is $29,340.53.
EOS

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1. Solve the DE y‘ – 2y/x = x².

Solution

An antiderivative of –2/x is –2 ln x. Then:

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2. Solve y‘ + y = 3e^x

Solution

An antiderivative of 1 is x, and so an integrating factor is e^x. We have:

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Solution

y‘ + (cot x) y – 2x – 1 = 0,
y‘ + (cot x) y = 2x + 1.

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4. Solve this initial-value problem:

Solution

(1 + sin x) y‘ + (cos x) y = sin² x,

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5. A tank contains 240 L of brine in which 6 kg of salt is dissolved. Then brine containing 0.06 kg of salt per litre from outside
    the tank flows into it at a rate of 24 L/min and the brine in it flows out of it at a rate of 12 L/min. The concentration of brine
    in the tank is kept uniform by stirring. Determine the amount of salt in the tank after 10 minutes.

Solution

Let S be the amount of salt in the tank at time t, where S is in kg and t in min. The initial time t = 0 is when the brine starts to
flow into and out of the tank. The rate of change of the amount of salt in the tank at time t is:

After 10 min, the amount of salt in the tank is 16 kg.

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6. The initial balance of a savings account was $2,000. Interest is paid into the account continuously at the variable rate of (1 +
    (t/60))% per month, where t is the number of months after the account was opened. Determine the amount in the account
    after 1 year.

Solution

Let a(t) be the amount in the account t months after the account was opened. The rate of increase of a(t) is:

2,000 = a(0) = Ce⁰ = C,

After 1 year or 12 months, the amount in the account is $2,282.22

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7. The equation y‘ + p(x)y = q(x)yⁿ, where n is a constant different from 0 and 1, is known as Bernoulli equation. (For n = 0,
    the equation becomes y‘ + p(x)y = q(x), and for n = 1, it becomes y‘ + (p(x) – q(x))y = 0. In both cases, the equation is
    of first-order and linear, and can be solved by the integrating-factor technique discussed in this section.)

a. Prove that the substitution v = y^1–n reduces the Bernoulli equation to a DE for v that is of first-order and linear.
b. Use the result of part a to solve the DE y‘ – 2xy = 5xy³.

Solution

a. Let v = y^1–n. Then:

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16.3.1 Equations With Constant Coefficients – Undetermined Coefficients

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1. Second-Order Linear Non-Homogeneous Differential Equations With     Constant Coefficients

In this section, we’ll use the abbreviations:

“DE”	for	“differential equation”,
“GS”	for	“general solution”,
“HDE”	for	“homogeneous differential equation”,
“HE”	for	“homogeneous equation”,
“NHDE”	for	“non-homogeneous differential equation”,
“NHE”	for	“non-homogeneous equation”, and
“PS”	for	“particular solution”.

Definition 1.1 – Second-Order Linear Non-Homogeneous Differential Equations With
                          Constant Coefficients

A differential equation of the form: y” + by‘ + cy = f (x), where b and c are constants and f is a continuous function that isn’t identically 0, is called a second-order linear non-homogeneous differential equation with constant coefficients.

Note that there are only the terms for y, y‘, and y” on the left-hand side. Everything else must be on the right-hand side and is
a part of f. Distinguish between the functions y and f. Observe that while the equation is linear, each function y, y‘, y”, and
f (x) doesn’t have to be linear. For example, y = 2x + 3 is linear while y = x² + 3 and y = e^2x + 3 aren’t.

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2. General Solutions Of The Non-Homogeneous Equations

Suppose y_n1 is a PS of the NHE:

y” + by‘ + cy = f (x)

and y_h is the GS of the corresponding HE:

y” + by‘ + cy = 0.

Then intuitively the sum y_h + y_n1 should be the GS of the NHE, because: (1) when we substitute y_h + y_n1 into the NHE, the
expression containing y_h and its derivatives equals 0, since y_h is the GS of the HE, (2) the expression containing y_n1 and its
derivatives equals f (x), since y_n1 is a PS of the NHE, and (3) there are 2 arbitrary constants in y_h + y_n1, they’re the arbitrary
constants of y_h. And so the substituted sum equals 0 + f (x) = f (x). This intuition is confirmed true in the theorem below.

Theorem 2.1 – General Solutions Of The Non-Homogeneous Equations

Let: ((GS Of NHE) = (GS Of HE) + (A PS Of NHE)).

Proof
First let’s prove that y_h + y_n1 is a solution of the NHE [2.1]. We have:

Thus y_n2 – y_n1 is a PS of the HE. Let y_h1 be that PS, so that y_n2 – y_n1 = y_h1, or y_n2 = y_h1 + y_n1. Now y_h1 can be obtained from y_h.
As a consequence, y_n2 can be obtained from y_h + y_n1. Hence y_h + y_n1 is the GS of the NHE [2.1].
EOP

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3. Principle Of Superposition

Because 0 + f (x) = f (x), by Theorem 2.1 we see that:

(GS of y” + by‘ + cy = 0 + f (x)) = (GS of y” + by‘ + cy = 0) + (PS of y” + by‘ + cy = f (x)).

This leads us to investigate to see if the property:

(solution of y” + by‘ + cy = f₁ (x) + f₂ (x)) = (solution of y” + by‘ + cy = f₁ (x)) + (solution of y” + by‘ + cy = f₂ (x))

is also true for any functions f₁ (x) and f₂ (x). Intuitively this is obviously true. Let y₁ be a solution of y” + by‘ + cy = f₁ (x) and
y₂ a solution of y” + by‘ + cy = f₂ (x). Substituting y₁ + y₂ for y in the expression y” + by‘ + cy, we get f₁ (x) + f₂ (x), because
the expression containing y₁ and its derivatives equals f₁ (x) and the expression containing y₂ and its derivatives equals f₂ (x).

It turns out that this property is indeed true, as asserted by Theorem 3.1 below, which establishes the relationship between the
solutions of the equations:

y” + by‘ + cy = f ₁(x),
y” + by‘ + cy = f ₂(x), and
y” + by‘ + cy = f ₁(x) + f ₂(x),

the relationship being that the sum of a solution of the 1st equation and a solution of the 2nd equation is a solution of the 3rd
equation. Letting “RHS” stands for “right-hand side”, this property can be thought of as:

(solution when RHS is) ( f₁(x) + f₂(x)) = (solution when RHS is) ( f₁(x)) + (solution when RHS is) ( f₂(x)).

This linearity property of the solutions is called the principle of superposition, as a solution of the 3rd equation is obtained by superimposing a solution of 1st equation on a solution of the 2nd equation.

Theorem 3.1 – Principle Of Superposition

If y1 is a solution of the equation: This property is called the principle of superposition, because a solution of Eq. [3.3] is obtained by superimposing a solution of Eq. [3.1] on a solution of Eq. [3.2].

Proof

So y₁ + y₂ is a solution of y” + by‘ +  cy = f₁(x) + f₂(x).

EOP

Remarks 3.1

a. The sum of a solution of Eq. [3.1] and a solution of Eq. [3.2] is a solution of Eq. [3.3], not a solution of the equation that’s
    the sum of Eqs. [3.1] and [3.2], which is 2( y” + by‘ + cy) = f₁(x) + f₂(x), or y” + by‘ + cy = (1/2)( f₁(x) + f₂(x)).

b. It’s straightforward to show that the principle of superposition is valid for any number of equations. Let n be any integer
    greater than or equal to 2. If y₁, y₂, …, and y_n are the solutions of the equations y” + by‘ +  cy = f₁(x), y” + by‘ +  cy =
    f₂(x), …, and y” + by‘ + cy = f_n(x) respectively, then y₁ + y₂ + … + y_n is a solution of the equation y” + by‘ +  cy = f₁(x) +
    f₂(x) + … + f_n(x).

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4. The Method Of Undetermined Coefficients

Clearly Theorem 2.1 tells us that to solve (find the GS of) a second-order linear NHDE with constant coefficients, we have to find the GS of the corresponding HE and a PS of the NHE, then we add them up to get the GS of the NHDE. We already know

how to find the GS of a HE. We’ll present a method to determine a PS of the NHE, called the method of undetermined
coefficients. The reason for this naming is clear later on. This method applies only to some types of functions on the right-hand

side of the NHE, ie the function f (x).

When f (x) Is A Non-0 Constant

Solving By Inspection

First we show that when f (x) is constant, we can find a PS of the NHE simply by inspection. Suppose we’re to find a PS of the
NHE:

y” – 7y‘ + 10y = 20.

Since a non-0 constant, 20 in this case, is a polynomial (of degree 0), finding a PS of the NHE where f (x) is a polynomial can
be done by the method of undetermined coefficients, as illustrated in Example 4.2 below.

Finding a PS of the NHE where f (x) is a non-0 constant can also be done by inspection. The function f (x) is the constant 20. If
y is a constant say k, then y‘ = 0 and y” = 0, so when we compute y” – 7y‘ + 10y, it’s a constant, namely 10k, and thus we
can choose k so that y” – 7y‘ + 10y = 20. We’ll then have 10k = 20, consequently k = 20/10 = 2. Hence we’ll try the function
y = 2 as a PS of the NHE. We use inspection in Example 4.1 below.

Example 4.1

Find the GS of the equation y” – 7y‘ + 10y = 20. Check the answer.

Solution
 

For a PS of the given NHE y” – 7y‘ + 10y = 20, let’s try y = 20/10 = 2. Then y‘ = 0, y” = 0, and y” – 7y‘ + 10y = 0 – 7(0)
+ 10(2) = 20. So a PS of the NHE is y = 2.

Thus the GS of the NHE is y = c₁e^5x + c₂e^2x + 2.

Check:

y = c₁e^5x + c₂e^2x + 2,
y‘ = 5c₁e^5x + 2c₂e^2x,
y” = 25c₁e^5x + 4c₂e^2x,
y” – 7y‘ + 10y = (25c₁e^5x + 4c₂e^2x) – 7(5c₁e^5x + 2c₂e^2x) + 10(c₁e^5x + c₂e^2x + 2) = 20.

Indeed the GS y = c₁e^5x + c₂e^2x + 2 is correct. 
EOS

Finding A Particular Solution By The Method Of Undetermined Coefficients

When f (x) Is A Polynomial Function

Let’s find a PS of the NHE:

y” + 2y‘ – 3y = 2x.

The function f (x) is a linear function, f (x) = 2x. So y should also be a linear function y = a₁x + a₀, so that when we compute
y” + 2y‘ – 3y, it’s also a linear function, and thus we can select the coefficients a₁ and a₀ that make y” + 2y‘ – 3y equal 2x.
Let’s try y = ax, where a is a constant to be determined such that y” + 2y‘ – 3y = 2x. We have:

y = ax,     y‘ = a,     y” = 0,
2x = y” + 2y‘ – 3y = 0 + 2a – 3ax = – 3ax + 2a,
– 3ax + 2a = 2x,
these 2 linear functions are the same thing if the coefficients of like powers of x are equal; so let’s equate the coefficients of
like powers of x:

a = –2/3 and a = 0, impossible,

the system has no solutions.

The trial PS y = ax doesn’t work.

So what do we do? Ok, there’s one remaining thing to do: it’s to try the linear function of the general form y = a₁x + a₀, where
a₁ and a₀ are constants to be determined such that y” + 2y‘ – 3y = 2x. We have:

y = a₁x + a₀,     y‘ = a₁,     y” = 0,

2x = y” + 2y‘ – 3y = 0 + 2a₁ – 3(a₁x + a₀) = – 3a₁x + 2a₁ – 3a₀,
– 3a₁x + 2a₁ – 3a₀ = 2x,
these 2 linear functions are the same thing if the coefficients of like powers of x are equal; so let’s equate the coefficients of
like powers of x:

solving this system we get a₁ = –2/3 and a₀ = 2(–2/3)/3 = –4/9.

The trial PS y = a₁x + a₀ works. A PS of the NHE is y = – (2/3)x – 4/9.

Example 4.2

Solve the NHE y” + 2y‘ – 3y = 2x by going thru the following steps:
a. Solve the HE y” + 2y‘ – 3y = 0.
b. Find a PS of the given NHE. Verify the answer.
c. Form the GS of the NHE. Verify the answer.

Solution

    Verification:

    Indeed the GS is correct.

EOS

The Method Of Undetermined Coefficients

In Example 4.2 part b, since 2x is a linear function, a PS y should also be a linear function, so that y‘ and y” also are, then y”
+ 2y‘ – 3y also is, thus we can make it equal the linear function 2x. Consequently, as a PS, we try the linear function of the
general form y = a₁x + a₀, not of the non-general form y = ax, as seen in the discussion preceding Example 4.2. We compute
y‘, y”, then y” + 2y‘ – 3y. For y” + 2y‘ – 3y to equal 2x, its x-expression must equal 2x. Hence we equate its x-expression
with 2x. For the 2 expressions to be the same thing, ie to be equal for all x, the coefficients of like powers of x must be equal.
It follows that we equate the coefficients of like powers of x (the constants are the coefficients of x⁰ = 1), get a system of 2
equations in a₁ and a₀, solve it, obtain a₁ = –2/3 and a₀ = –4/9. Therefore the PS is y = – (2/3)x – 4/9.

We try the trial solution y = a₁x + a₀, where the coefficients a₁ and a₀ are to be determined. Of course initially they’re
undetermined. Hence this method of trying a PS with initially-undetermined coefficients is called the method of
undetermined coefficients.

As asked, we verify the PS and the GS of the NHE, and find that indeed they’re correct. The procedure to determine a PS
assures that the PS must be correct, and thus, by Theorem 2.1, the GS must also be correct. Example 4.2 asks to verify simply because at the first contact with the method, it’s good to assure you clearly that of course the method works. In solutions to

problems, if you’re not asked to verify, then you don’t have to verify.

Example 4.3

Solve y” – y = x².

Note

The function f (x) is a quadratic function, f(x) = x². So y should also be a quadratic function, so that when we compute y” – y,
it’s also a quadratic function and thus has a chance to equal x². This convinces us, in using the method of undetermined
coefficients, to try as a PS of the given NHE the quadratic function of the general form y = a₂x² + a₁x + a₀, where a₂, a₁, and a₀
are constants to be determined in such a way that y” – y = x². As in the case for f (x) being a linear function, we must try the
general form of the polynomial, in this case the general form of the quadratic function.

Solution
 

a₂ = –1,     a₁ = 0,     a₀ = 2a₂ = 2(–1) = –2,
y = – x² – 2.

The GE of the NHE is y = c₁e^x + c₂e^–x – x² – 2.
EOS

When f (x) Is An Exponential Function

Let’s find the working trial solution of the NHE:

y” – 3y‘ + y = 7e^2x.

The function f (x) is an exponential function with exponent 2x, f (x) = 7e^2x. So y should also be an exponential function with
the same exponent 2x, so that when we compute y” – 3y‘ + y, it’s also an exponential function with the same exponent 2x, as
the derivative of any order of e^2x is a constant multiple of e^2x, and thus y” – 3y‘ + y can be made to equal 7e^2x. The trial
solution is the general form of the exponential function e^2x, which is y = ae^2x. This trial solution works, as shown in Example
4.4 below. Note that the trial solution when f (x) = ke^sx, where k and s are constants, is ae^sx, not ake^sx.

Example 4.4

Solve the DE y” – 3y‘ + y = 7e^2x.

Solution

EOS

When f (x) Is A Sine Or Cosine Function

Let’s determine the working trial solution of the NHE:

y” + 5y‘ + 4y = 2 sin x.

The function f (x) is a sine function, f (x) = 2 sin x. So the x-expression of y” + 5y‘ + 4y must be able to equal 2 sin x. This
suggests that we should use a sine function as a trial solution. The derivative of any order of sin x is a constant multiple of
either sin x or cos x. This re-enforces the suggestion to use a sine function. Thus let’s try y = a sin x, where a is a constant to
be determined so that y” + 5y‘ + 4y = 2 sin x. We have:

y = a sin x,     y‘ = a cos x,     y” = – a sin x,
2 sin x = y” + 5y‘ + 4y = – a sin x + 5a cos x + 4a sin x = 3a sin x + 5a cos x,
3a sin x + 5a cos x = 2 sin x,
equate the coefficients of like trigonometric functions of x:

this system has no solutions.

So the trial solution y = a sin x doesn’t work. The reason is discussed after the Solution of Example 4.5 below.

So what do we do? Ok, there’s one remaining thing to do: it’s to try the cosine+sine function of the general form y = A cos x +
B sin x, where A and B are constants to be determined so that y” + 5y‘ + 4y = 2 sin x. This trial PS works, as seen in
Example 4.5 below. The reason is discussed after the Solution of Example 4.5.

Example 4.5

Solve y” + 5y‘ + 4y = 2 sin x.

Solution

EOS

Why y = a sin x Doesn’t Work But y = A cos x + B sin x Does

For the trial solution y = ax, the x-expression of y” + 5y‘ + 4y is 3a sin x + 5a cos x, where the coefficients of cos x and
sin x are products containing the same factor a. When we remove the term 5a cos x by setting a = 0, the term 3a sin x will
also disappear. That’s the reason why y = a sin x doesn’t work.

For the trial solution y = A cos x + B sin x, the x-expression of y” + 5y‘ + 4y is (– 5A + 3B) sin x + (3A + 5B) cos x, where
the coefficients of cos x and sin x aren’t products containing the same factor that has to be set to 0. When we remove the
term (3A + 5B) cos x by setting 3A + 5B = 0, the term (– 5A + 3B) sin x won’t disappear. That’s the reason why y = A cos x
+ B sin x works.

Example 4.6

Find the general solution of the DE y” – 3y‘ + 2y = 2 cos 4x.

Note

Similar to the discussions surrounding Example 4.5. The general form of the cosine+sine function cos 4x + sin 4x is A cos 4x
+ B sin 4x.

Solution

EOS

Eligible Functions

In this section, eligible functions refer to functions that are eligible for the method of undetermined coefficients. They’re
represented by the function f (x) in the NHE:

y” + by‘ + cy = f (x).

They’re the polynomial, sine, cosine, and exponential functions.

If f (x) is a polynomial function of degree n: derivatives of orders 0 thru to 2, which appear on the left-hand side of the
equation, of the polynomial function f (x) of degree n are polynomial functions of degree n and less; so the trial PS of the NHE
is a polynomial function of degree n of the general form:

y = a_nxⁿ + a_n–1x^n–1 + a_n–2x^n–2 + … + a₂x² + a₁x + a₀,

so that the two sides can be made equal by equating the coefficients of like powers of x.

If f (x) is a sine or cosine function: derivatives of the sine function f (x) = r sin sx and of the cosine function f (x) = r cos sx
are each constant multiples of the sine and cosine functions sin sx and cos sx; so the trial PS is the cosine+sine function of the
general form:

y = A cos sx + B sin sx.

If f (x) is an exponential function: derivatives of the exponential function f (x) = re^sx are constant multiples of the exponential
function e^sx; so the trial PS is the exponential of the general form:

y = ae^sx.

For non-eligible functions, consider the following example:

If f (x) = ln x, letting the trial solution be y = a ln x would lead to y” + by‘ + cy containing a constant multiple of –1/x², and
thus y” + by‘ + cy can’t be made equal to ln x.

There are situations where the trial PSs have to be modified. This modification is discussed in the next section, where part 2 of this topic is presented. Thus if you’re encountering this topic for the first time, you’re reminded that your knowledge of it will be

complete only after you successfully finish its part 2.

When f (x) Is A Sum Of Eligible Functions

Example 4.7

Consider the equation y” – y = x² + 3 cos x.
a. Solve the equation y” – y = x².
b. Solve the equation y” – y = 3 cos x.
c. Solve the equation y” – y = x² + 3 cos x using the principle of superposition.
d. Solve the equation y” – y = x² + 3 cos x using the method of undetermined coefficients.

Note

For part d. The function f (x) is the sum of a quadratic function and a cosine function, f (x) = x² + 3 cos x. So naturally, as a
PS of the NHE y” – y = x² + 3 cos x, we’ll try the sum of the general quadratic function and the general form of the
cosine+sine function, that sum being y = a₂x² + a₁x + a₀ + A cos x + B sin x, so that when we compute y” – y, it’ll contain x²
and cos x, and thus it’ll be possible to make it equal x² + 3 cos x by the process of equating coefficients.

Solution

    a₂ = –1, a₁ = 0, a₀ = 2(–1) = –2,
    a PS of the NHE is y = – x² – 2.

    So the required GS is:

    y = c₁e^x + c₂e^–x – x² – 2.

b. By part a, the GS of the HE y” – y = 0 is y = c₁e^x + c₂e^–x.

    For a PS of the NHE y” – y = 3 cos x, let y = A cos x + B sin x. We have:

    y = A cos x + B sin x,     y‘ = – A sin x + B cos x,     y” = – A cos x – B sin x,

    3 cos x = y” – y = (– A cos x – B sin x) – (A cos x + B sin x) = – 2A cos x – 2B sin x,
    – 2A cos x – 2B sin x = 3 cos x,

d. By part a, the GS of the HE y” – y = 0 is y = c₁e^x + c₂e^–x.

    For a PS of the NHE y” – y = x² + 3 cos x, let y = a₂x² + a₁x + a₀ + A cos x + B sin x. We have:

    y = a₂x² + a₁x + a₀ + A cos x + B sin x,     y‘ = 2a₂x + a₁ – A sin x + B cos x,     y” = 2a₂ – A cos x – B sin x,

EOS

Using The Principle Of Superposition

As Example 4.7 part c shows, when f (x) is the sum of two or more eligible functions, we can use the principle of
superposition, as shown in Theorem 3.1. A PS of:

y” + by‘ + cy = f₁(x) + f₂(x) + … f_n(x)

equals the sum of PSs of:

y” + by‘ + cy = f₁(x),     y” + by‘ + cy = f₂(x),     …,     and     y” + by‘ + cy = f_n(x).

When f (x) Is A Product Of A Polynomial Function And An Exponential Function

Let’s find a PS of the NHE:

y” + y = xe^2x.

The function f (x) is the product of a polynomial function of degree 1 and an exponential function, f (x) = xe^2x. So naturally, as a PS of this equation, we’ll try the product of the general polynomial function of degree 1 and the general form of the

exponential function e^2x, that product being y = (a₁x + a₀)ae^2x, so that when we compute y” + y, it’ll contain the product xe^2x,

and thus it’ll be possible to make it equal xe^2x by the process of equating coefficients. We have:

y = (a₁x + a₀)ae^2x,
y‘ = aa₁e^2x + (2aa₁x + 2aa₀)e^2x,
y” = 2aa₁e^2x + 2aa₁e^2x + (4aa₁x + 4aa₀)e^2x = 4aa₁e^2x + (4aa₁x + 4aa₀)e^2x,

xe^2x = y” + y = 4aa₁e^2x + (4aa₁x + 4aa₀)e^2x + (a₁x + a₀)ae^2x = (5aa₁x  + 4aa₁ + 5aa₀)e^2x,
(5aa₁x  + 4aa₁ + 5aa₀)e^2x = xe^2x,
5aa₁x  + 4aa₁ + 5aa₀ = x,

a₁ = 1/(5a), a₀ = – 4a(1/(5a)) / (5a) = – 4/(25a),
letting a = 1 we get a₁ = 1/5 and a₀ = –4/25, and a PS is y = ((1/5)x – (4/25))e^2x.

Now (a₁x + a₀)(ae^2x) = (aa₁x + aa₀)e^2x, and aa₁ and aa₀ are constants. Thus we can simply use y = (A₁x + A₀)e^2x, where A₁
and A₀ are constants, as the trial PS. Re-utilizing the letters a₁ and a₀ in place of A₁ and A₀ respectively, our trial PS is y = (a₁x
+ a₀)e^2x. This is the same as using a = 1. It works all right, giving the same PS, as seen in Example 4.8 below. When a
constant can be combined with all the remaining constants, it can be removed.

Example 4.8

Consider the equation y” + y = xe^2x.
a. Solve the equation y” + y = x.
b. Solve the equation y” + y = e^2x.
c. Solve the equation y” + y = xe^2x.
d. Does the GS of y” + y = xe^2x equal the product of the GS of y” + y = x and the GS of y” + y = e^2x?

Solution

    a₁ = 1, a₀ = 0,
    a PS of the NHE is y = x.

    So the GS of y” + y = x is:

    y = x + c₁ cos x + c₂ sin x.

b. By part a, the GS of the HE y” + y = 0 is y = c₁ cos x + c₂ sin x.

    For a PS of the NHE y” + y = e^2x, let y = ae^2x. We have:

     y = ae^2x,     y‘ = 2ae^2x,     y” = 4ae^2x,

    e^2x = y” + y = 4ae^2x + ae^2x = 5ae^2x,
    5ae^2x = e^2x,
    5a = 1,
    a = 1/5,

    a PS of the NHE is y = (1/5)e^2x.

    So the GS of y” + y = e^2x is:

    a₁ = 1/5,
    a₀ = – 4(1/5)/5 = –4/25,
    a PS of the NHE is y = ((1/5)x – (4/25))e^2xy” + y = xe^2x is:

    The product [1] x [2] contains the term x(c₁ cos x) = c₁x cos x, which [3] doesn’t contain. So the product [1] x [2] can’t
    equal [3]. Thus the answer is no.

EOS

When f (x) Is A Product Of A Polynomial Function And A Sine Or Cosine Function

Let’s determine the working trial PS of the NHDE:

y” – y = x sin x.

Let’s try y = (a₁x + a₀)(A cos x + B sin x). We have:

y = (a₁x + a₀)(A cos x + B sin x),
y‘ = a₁(A cos x + B sin x) + (a₁x + a₀)(– A sin x + B cos x),
y” = a₁(– A sin x + B cos x) + a₁(– A sin x + B cos x) + (a₁x + a₀)(– A cos x – B sin x)
     = 2a₁(– A sin x + B cos x) – (a₁x + a₀)(A cos x + B sin x),
x sin x = y” – y
           = 2a₁(– A sin x + B cos x) – (a₁x + a₀)(A cos x + B sin x) – (a₁x + a₀)(A cos x + B sin x)
           = 2a₁(– A sin x + B cos x) – 2(a₁x + a₀)(A cos x + B sin x),
           = (– 2a₁A – 2a₁xB – 2a₀B) sin x + (2a₁B – 2a₁xA – 2a₀A) cos x
           = (– 2a₁Bx – 2a₁A – 2a₀B) sin x + (– 2a₁xA + 2a₁B – 2a₀A) cos x
(– 2a₁Bx – 2a₁A – 2a₀B) sin x + (– 2a₁Ax + 2a₁B – 2a₀A) cos x = x sin x,

as [1] and [3] show, the system has no solution, the trial solution y = (a₁x + a₀)(cos x + sin x) fails too.

In Example 4.8, we have f (x) = xe^2x, and the working trial solution is y = (a₁x + a₀)e^2x, obtained by multiplying the coefficient
of e^2x into the polynomial, so that e^2x has a coefficient of 1. Let’s try the same strategy: move A and B from A cos x + B sin x
to the polynomial, so that cos x and sin x each have a coefficient of 1. We have:

y = (a₁x + a₀)(A cos x + B sin x) = (a₁x + a₀)A cos x + (a₁x + a₀) B sin x = (a₁Ax + a₀A) cos x + (a₁Bx + a₀B) sin x.

Let’s re-use the letters a₁ and a₀ for the products a₁A and a₀A respectively, and let b₁ = a₁B and b₀ = a₀B. Then y =
(a₁x + a₀) cos x + (b₁x + b₀) sin x. That’s what we want! Let’s try it: y = (a₁x + a₀) cos x + (b₁x + b₀) sin x, done in
Example 4.9 below. It works, as seen in the Solution! Wow!

Example 4.9

Solve the NHDE y” – y = x sin x.

Solution

For a PS of the given NHE y” – y = x sin x, let y = (a₁x + a₀) cos x + (b₁x + b₀) sin x. We have:

y = (a₁x + a₀) cos x + (b₁x + b₀) sin x,
y‘ = a₁ cos x – (a₁x + a₀) sin x + b₁ sin x + (b₁x + b₀) cos x,
y” = – a₁ sin x – a₁ sin x – (a₁x + a₀) cos x + b₁ cos x + b₁ cos x – (b₁x + b₀) sin x
    = (– a₁x – a₀ + 2b₁) cos x + (– b₁x – 2a₁ – b₀) sin x,
x sin x = y” – y
           = ((– a₁x – a₀ + 2b₁) cos x + (– b₁x – 2a₁ – b₀) sin x) – ((a₁x + a₀) cos x + (b₁x + b₀) sin x)
           = (– 2b₁x – 2a₁ – 2b₀) sin x + (– 2a₁x – 2a₀ + 2b₁) cos x,
(– 2b₁x – 2a₁ – 2b₀) sin x + (– 2a₁x – 2a₀ + 2b₁) cos x = x sin x,

b₁ = –1/2, a₁ = 0, b₀ = 0, a₀ = – 2(–1/2) / (–2) = –1/2,
a PS of the NHE is y = – (1/2) cos x – (1/2)x sin x, or y = – (1/2)(cos x + x sin x).

Thus the GS of the given NHE is y = c₁e^x + c₂e^–x – (1/2)(cos x + x sin x).

EOS

When f (x) Is A Product Of An Exponential Function And A Sine Or Cosine Function

Let’s find a PS of the NHE:

y” – 3y‘ + 2y = e^x sin x.

Let y = ae^x(A cos x + B sin x). Then y = e^x(aA cos x + aB sin x). So we can simply utilize y = e^x(A cos x + B sin x). It
works ok, as evidenced in Example 4.10 below.

Example 4.10

Determine the GS of y” – 3y‘ + 2y = e^x sin x.

Solution

EOS

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5. Modification Of The Method Of Undetermined Coefficients

When f (x) Is A Polynomial

Let’s find a PS of the NHE:

y” + y‘ = x³ – x².

Let’s try y = a₃x³ + a₂x² + a₁x + a₀. We have:

y = a₃x³ + a₂x² + a₁x + a₀,     y‘ = 3a₃x² + 2a₂x + a₁,     y” = 6a₃x + 2a₂,

x³ – x² = y” + y‘ = (6a₃x + 2a₂) + (3a₃x² + 2a₂x + a₁) = 3a₃x² + (6a₃ + 2a₂)x + 2a₂ + a₁,
3a₃x² + (6a₃ + 2a₂)x + 2a₂ + a₁ = x³ – x².

We may add the term 0x³ to the left-hand side, but then we can’t equate 0 to the coefficient 1 of x³ on the right-hand side, and so the process of equating coefficients can’t be done.

The trial solution y = a₃x³ + a₂x² + a₁x + a₀ doesn’t work. That’s because when we compute (the x-expressions of ) y‘ and y”,
they’re polynomials of degree 2, and so is y” + y‘, while x³ – x² is a polynomial of degree 3. The term containing y, which is of
degree 3 in x, is absent from the left-hand side.

So what do we do? Ok, let’s try a polynomial of degree 4, so that when we compute y” + y‘, it’ll be a polynomial of degree 3,
the same as that of x³ – x². Let y = a₄x⁴ + a₃x³ + a₂x² + a₁x + a₀. We have:

y = a₄x⁴ + a₃x³ + a₂x² + a₁x + a₀,     y‘ = 4a₄x³ + 3a₃x² + 2a₂x + a₁,     y” = 12a₄x² + 6a₃x + 2a₂,

x³ – x² = y” + y‘ = (12a₄x² + 6a₃x + 2a₂) + (4a₄x³ + 3a₃x² + 2a₂x + a₁) = 4a₄x³ + (12a₄ + 3a₃)x² + (6a₃ + 2a₂)x + 2a₂ + a₁,
4a₄x³ + (12a₄ + 3a₃)x² + (6a₃ + 2a₂)x + 2a₂ + a₁ = x³ – x².

The constant term a₀ is absent from the left-hand side, and thus can’t be determined. The trial solution y = a₄x⁴ + a₃x³ + a₂x²
+ a₁x + a₀ doesn’t work either.

So, again, what do we do? Ok, let’s multiply the initial trial solution a₃x³ + a₂x² + a₁x + a₀ by x to get the modified trial solution
y = a₃x⁴ + a₂x³ + a₁x² + a₀x. This trial PS works, as will be seen in Example 5.1 below.

When f (x) is a polynomial and y is absent from the left-hand side (because its coefficient is 0), the modified trial solution is a
polynomial of degree greater than that of the polynomial f (x) by 1 and with the constant coefficient equal to 0. In the term
a_ixⁱ⁺¹, the subscript i of the coefficient a_i is less than the exponent i + 1 of xⁱ⁺¹ by 1, to reveal the multiplication by x and to
stop at a₀x where the subscript is already the lowest 0. This is an example where the method of undetermined coefficients
must be modified.

Example 5.1

Find the GS of the DE y” + y‘ = x³ – x².

Solution

a₃ = 1/4,     a₂ = (– 1 – 12(1/4)) / 3 = –4/3,     a₁ = –6(–4/3)/2 = 4,     a₀ = –2(4) = –8,
a PS of the NHE is y = (1/4)x⁴ – (4/3)x³ + 4x² – 8x.

So the GS of the given NHE is y = c₁ + c₂e^–x + (1/4)x⁴ – (4/3)x³ + 4x² – 8x.
EOS

When The Initial Trial PS Of The NHE Is A PS Of The HE

Multiplying By x

Let’s determine a PS of the NHE:

y” – y = 2e^x.

Let’s try y = ae^x. We have:

y = ae^x,     y‘ = ae^x,     y” = ae^x,

2e^x = y” – y = ae^x – ae^x = 0.

The case of the polynomial in Example 5.1 suggests that we can try the multiplication of the initial trial solution ae^x by x. Let’s
try y = axe^x. This trial solution works quite well, as shown in Example 5.2 below. Note that axe^x isn’t a PS of the HE for any
values of c₁ and c₂.

In the solution we of course first find the GS of the HE. If we observe that the initial trial PS g(x) of the NHE is a PS of the HE,
then we multiply it by x, if xg(x) obtained isn’t a PS of the HE, then we use it as the modified trial PS. This is another example
where the method of undetermined coefficients must be modified.

Example 5.2

Solve y” – y = 2e^x.

Solution

For a PS of the given NHE y” – y = 2e^x. The function ae^x is a PS of the HE and the function axe^x isn’t. So let’s try y = axe^x.
We have:

y = axe^x,     y‘ = ae^x + axe^x,     y” = ae^x + (ae^x + axe^x) = 2ae^x + axe^x,

2e^x = y” – y = (2ae^x + axe^x) – axe^x = 2ae^x,
2ae^x = 2e^x,
a = 1,
a PS is y = xe^x.

So the GS of the given NHE is y = c₁e^x + c₂e^–x + xe^x, or y = (x + c₁)e^x + c₂e^–x.
EOS

Example 5.3

Solve the initial-value problem y” + y = cos x, y(0) = 2, y‘(0) = –3.

Note

For a PS of the given NHE. Let’s try y = A cos x + B sin x. We have:

y = A cos x + B sin x,     y‘ = – A sin x + B cos x,     y” = – A cos x – B sin x,

cos x = y” + y = (– A cos x – B sin x) + (A cos x + B sin x) = 0.

Solution

c₁ = 2,

c₂ = –3.

Thus the unique solution of the given initial-value problem is y = 2 cos x – 3 sin x + (1/2)x sin x.

EOS

Multiplying By xⁿ

Let’s find a PS of the equation:

y” – 2y‘ + y = e^x.

Let’s try y = ae^x. We have:

y = ae^x,     y‘ = ae^x,     y” = ae^x,

e^x = y” – 2y‘ + y = ae^x – 2ae^x + ae^x = 0.

The trial PS y = ae^x fails. It’s a PS of the HE y” – 2y‘ + y = 0, and thus can’t be a PS of the NHE. Let’s determine the GS of the
HE. We have (–2)² – 4(1) = 0, –b/2 = –(–2)/2 = 1, and the GS is y = c₁e^x + c₂xe^x. The trial PS y = ae^x of the NHE is a PS of
the HE, obtained from the GS of the HE by letting c₁ = a and c₂ = 0.

Now let’s multiply the initial trial PS ae^x by x to get axe^x. But axe^x is still a PS of the HE, obtained by letting c₁ = 0 and c₂ = a.
So y = axe^x would fail too. Let’s check:

y = axe^x,     y‘ = ae^x + axe^x,     y” = ae^x + ae^x + axe^x = 2ae^x + axe^x,

e^x = y” – 2y‘ + y = 2ae^x + axe^x – 2(ae^x + axe^x) + axe^x = 0.

Indeed y = axe^x fails too.

So what do we do? Ok, let’s multiply the initial trial PS ae^x by x² to get ax²e^x. Now ax²e^x isn’t a PS of the HE. Thus let’s try y =
ax²e^x. This trial solution works, as evidenced in Example 5.4 below. Note that ax²e^x isn’t a PS of the HE for any values of c₁
and c₂. This is another example where the method of undetermined coefficients must be modified.

In the solution we of course first find the GS of the HE. If we observe that the initial trial PS g(x) of the NHE is a PS of the HE,
then we multiply it by xⁿ, where n is the smallest positive integer such that xⁿg(x) isn’t a PS of the HE, and we use the
modified trial PS y = xⁿg(x).

Example 5.4

Determine the GS of the DE y” – 2y‘ + y = e^x.

Solution
For the GS of the HE y” – 2y‘ + y = 0, we have (–2)² – 4(1) = 0, –b/2 = –(–2)/2 = 1, and the GS is y = c₁e^x + c₂xe^x.

For a PS of the given NHE y” – 2y‘ + y = e^x. The functions y = ae^x and axe^x are PSs of the HE. So let’s try y = ax²e^x. We
have:

y = ax²e^x,     y‘ = 2axe^x + ax²e^x,     y” = (2ae^x + 2axe^x) + (2axe^x + ax²e^x) = 2ae^x + 4axe^x + ax²e^x,

e^x = y” – 2y‘ + y = (2ae^x + 4axe^x + ax²e^x) – 2(2axe^x + ax²e^x) + ax²e^x = 2ae^x,
2ae^x = e^x,
2a = 1,
a = 1/2,
a PS of the NHE is y = (1/2)x²e^x.

Thus the GS of the given NHE is y = c₁e^x + c₂xe^x + (1/2)x²e^x, or y = ((1/2)x² + c₂x + c₁)e^x.

EOS

When One Or More Terms Of The Initial Trial PS Of The NHE Is A PS Of The HE

Let’s determine a PS of the equation:

y” + y = x sin x.

Let’s try y = (a₁x + a₀) cos x + (b₁x + b₀) sin x. We have:

y = (a₁x + a₀) cos x + (b₁x + b₀) sin x,
y‘ = a₁ cos x – (a₁x + a₀) sin x + b₁ sin x + (b₁x + b₀) cos x = (b₁x + a₁ + b₀) cos x + (– a₁x – a₀ + b₁) sin x,

y” = b₁ cos x – (b₁x + a₁ + b₀) sin x – a₁ sin x + (– a₁x – a₀ + b₁) cos x = (– a₁x – a₀ + 2b₁) cos x + (– b₁x – 2a₁ – b₀) sin x,

x sin x = y” + y = 2b₁ cos x – 2a₁ sin x,
2b₁ cos x – 2a₁ sin x = x sin x.

Now the factor x in x sin x is a polynomial of degree 1, but there’s no polynomial of degree 1 as a factor in – 2a₁ sin x. The
process of equating coefficients can’t be done. The trial PS y = (a₁x + a₀) cos x + (b₁x + b₀) sin x doesn’t work. Let’s find the
GS of the HE y” + y = 0. As done in Example 5.3, it’s y = c₁ cos x + c₂ sin x. The trial PS is:

y = (a₁x + a₀) cos x + (b₁x + b₀) sin x = a₁x cos x + a₀ cos x + b₁x sin x + b₀ sin x.

Clearly the terms a₀ cos x and b₀ sin x are solutions of the HE. They disappear from the x-expression of y” + y. So the trial
solution is the same as if it’s y = a₁x cos x + b₁x sin x. Let’s check:

y = a₁x cos x + b₁x sin x,
y‘ = a₁ cos x – a₁x sin x + b₁ sin x + b₁x cos x = (b₁x + a₁) cos x + (– a₁x + b₁) sin x,
y” = b₁ cos x – (b₁x + a₁) sin x – a₁ sin x + (– a₁x + b₁) cos x = (– a₁x + 2b₁) cos x + (– b₁x – 2a₁) sin x,

y” + y = 2b₁ cos x – 2a₁ sin x,

which is the same as found above. Indeed the trial solution is the same as if it’s y = a₁x cos x + b₁x sin x, which is
incomplete, and consequently doesn’t work.

Let’s multiply the initial trial solution by x to get the modified trial solution (a₁x² + a₀x) cos x + (b₁x² + b₀x) sin x. No term of
this modified trial solution is a solution of the HE. This modified trial solution works quite well, as shown in Example 5.5 below.

In the solution we of course first find the GS of the HE. If we observe that one or more terms of the initial trial PS g(x) of the
NHE are PSs of the HE, then we multiply g(x) by xⁿ, where n is the smallest positive integer such that no term of xⁿg(x) is a
solution of the HE, and we use the modified trial PS y = xⁿg(x).

Example 5.5

Solve the equation y” + y = x sin x.

Solution

b₁ = 0,     a₁ = –1/4,     b₀ = –2(–1/4)/2 = 1/4,     a₀ = 0,
a PS of the NHE is y = (–1/4)x² cos x + (1/4)x sin x.

So the GS of the given NHE is y = c₁ cos x + c₂ sin x + (–1/4)x² cos x + (1/4)x sin x, or y = (c₁ – (1/4)x²) cos x +
(c₂ + (1/4)x) sin x.

EOS

Terms And Factors

Let’s recall the following terminology. In the expression:

y = ax²e^x + bx sin 3x,

ax²e^x and bx sin 3x are terms of y. In the expression:

y = ax²e^x,

a, x² (or x, as x² = xx), and e^x are factors of y.

In Example 5.5, the modified trial solution of the NHE is:

y = (a₁x² + a₀x) cos x + (b₁x² + b₀x) sin x = x²(a₁ cos x + b₁ sin x) + x(a₀ cos x + b₀ sin x).

The items a₁ cos x, b₁ sin x, a₀ cos x, and b₀ sin x are solutions of the HE, but they cause no problem, because they’re not
terms of y. (The terms are a₁x² cos x, b₁x² sin x, a₀x cos x, and b₀x sin x.)

In Example 5.2, the initial trial solution is y = ae^x, which is a solution of the HE. The modified trial solution is y = axe^x =
x(ae^x), where the factor ae^x is a solution of the HE, but it causes no problem, because it’s not a term of y. Similarly for
Examples 5.3 and 5.4.

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We now recap the results of this section. Consider the NHE:

y” + by‘ + cy = f (x)

and its associated HE:

y” + by‘ + cy = 0.

Case 1. No part of the trial PS of the NHE is a solution of the HE. A PS of the NHE has the form of y as described in the table
below, where P_n(x) and Q_n(x) are polynomials of degree n.

^{6.1} Examples 4.2 and 4.3

^{6.2} Example 5.1

^{6.3} Example 4.4
^{6.4} Example 4.5

^{6.5} Example 4.6
^{6.6} Example 4.8
^{6.7} Example 4.9
^{6.8} Example 4.10

Case 2. One or more terms of the initial trial PS of the NHE are solutions of the HE. Multiply the appropriate trial solution y in
the table above by xⁿ, where n is the smallest positive integer such that no term of the obtained product is a solution of the HE.

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1. Find the GS of y” + 4y‘ + 4y = 12.

Solution

For the GS of the HE y” + 4y‘ + 4y = 0, we have 4² – 4(4) = 0, –b/2 = –4/2 = –2, and the GS is y = c₁e^–2x + c₂xe^–2x.

For a PS of the given NHE y” + 4y‘ + 4y = 12, let’s try y = 12/4 = 3. Then y‘ = 0, y” = 0, and y” + 4y‘ + 4y = 0 + 4(0) + 4(3) =
12. So a PS of the NHE is y = 3.

Thus the GS of the NHE is y = c₁e^–2x + c₂xe^–2x + 3.

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2. Determine the GS of y” – 3y‘ + 2y = x + 1.

Solution

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3. Find the unique solution of the initial-value problem:

Solution

a₂ = 1,     a₁ = 1,     a₀ = 1 – 2(1) = –1,
a PS is y = 1x² + 1x – 1 or y = x² + x – 1.

So the GS of the NHE is y = c₁ cos x + c₂ sin x + x² + x – 1. Then y‘ = – c₁ sin x + c₂ cos x + 2x + 1. We have:

3 = y(0) = c₁ cos 0 + c₂ sin 0 + 0² + 0 – 1 = c₁ – 1, c₁ = 4,

1 = y‘(0) = – c₁ sin 0 + c₂ cos 0 + 2(0) + 1 = c₂ + 1, c₂ = 0.

Thus the required solution is y = 4 cos x + 0 sin x + x² + x – 1, or y = 4 cos x + x² + x – 1.

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4. Solve y” – 3y‘ + 2y = 6e^3x.

Solution

For a PS of the given NHE y” – 3y‘ + 2y = 6e^3x, let y = ae^3x. We have:

y = ae^3x,     y‘ = 3ae^3x,     y” = 9ae^3x,

6e^3x = y” – 3y‘ + 2y = 9ae^3x – 3(3ae^3x) + 2ae^3x = 2ae^3x,
2ae^3x = 6e^3x,
2a = 6,
a = 3,
a PS is y = 3e^3x.

So the required GS is y = c₁e^2x + c₂e^x + 3e^3x.

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5. Find the GS of the equation y” + 4y = 3 sin x.

Solution

A = 0, B = 1,
a PS is y = 0 cos x + 1 sin x or y = sin x.

So the GS of the given NHE is y = c₁ cos 2x + c₂ sin 2x + sin x.

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6. Determine the GS of the DE y” – 2y‘ = 2 sin x + 2e^3x. Hint: use the principle of superposition.

Solution

For a PS of the NHE y” – 2y‘ = 2 sin x, let y = A cos x + B sin x. We have:

y = A cos x + B sin x,     y‘ = – A sin x + B cos x,     y” = – A cos x – B sin x,
2 sin x = y” – 2y‘ = (– A cos x – B sin x)­ ­– 2(– A sin x + B cos x) = (2A – B) sin x + (– A – 2B) cos x,
(2A – B) sin x + (– A – 2B) cos x = 2 sin x,

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7. Solve the initial-value problem y” – 4y‘ + 4y = 6xe^3x, y(0) = 0, y‘(0) = 3.

Solution

For the GS of the HE y” – 4y‘ + 4y = 0, we have (–4)² – 4(4) = 0, –b/2 = –(–4)/2 = 2, and the GS is y = c₁ e^2x + c₂xe^2x.

For a PS of the given NHE, let y = (a₁x + a₀)e^3x. We have:

y = (a₁x + a₀)e^3x,
y‘ = a₁e^3x + 3(a₁x + a₀)e^3x = (a₁ + 3a₀)e^3x + 3a₁xe^3x,
y” = 3(a₁ + 3a₀)e^3x + 3a₁e^3x + 9a₁xe^3x = (6a₁ + 9a₀)e^3x + 9a₁xe^3x,
6xe^3x = y” – 4y‘ + 4y = (6a₁ + 9a₀)e^3x + 9a₁xe^3x – 4((a₁ + 3a₀)e^3x + 3a₁xe^3x) + 4(a₁x + a₀)e^3x = (a₁x + 2a₁ + a₀)e^3x,
(a₁x + 2a₁ + a₀)e^3x = 6xe^3x,

a₁x + 2a₁ + a₀ = 6x,

a₁ = 6,     a₀ = –2(6) = –12,
a PS of the NHE is y = (6x – 12)e^3x, or y = 6(x – 2)e^3x.

So the GS of the NHE is y = c₁e^2x + c₂xe^2x + 6(x – 2)e^3x. We have:

0 = y(0) = c₁e⁰ + 0 + 6(0 – 2)e⁰ = c₁ – 12,     c₁ = 12,
y‘ = 2c₁e^2x + c₂e^2x + 2c₂xe^2x + 6e^3x + 18(x – 2)e^3x = (2c₁ + c₂)e^2x + 2c₂xe^2x + (18x – 30)e^3x,
3 = y‘(0) = (2(12) + c₂)e⁰ + 0 + (0 – 30)e⁰ = c₂ – 6,     c₂ = 9.

Thus the unique solution of the given initial-value problem is y = 12e^2x + 9xe^2x + 6(x – 2)e^3x.

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8. Solve the NHE y” + 4y = 16x sin 3x.

Solution

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9. Find the GS of the NHDE y” + 4y‘ + 4y = e^2x cos x.

Solution

For the GS of the HE y” + 4y‘ + 4y = 0, we have 4² – 4(4) = 0, –b/2 = –4/2 = –2, and the GS is y = c₁e^–2x + c₂xe^–2x.

For a PS of the given NHE, let y = e^2x(A cos x + B sin x). We have:

y = e^2x(A cos x + B sin x),
y‘ = 2e^2x(A cos x + B sin x) + e^2x(– A sin x + B cos x),
y” = 4e^2x(A cos x + B sin x) + 2e^2x(– A sin x + B cos x) + 2e^2x(– A sin x + B cos x) + e^2x(– A cos x – B sin x)
     = 4e^2x(A cos x + B sin x) + 4e^2x(– A sin x + B cos x) – e^2x(A cos x + B sin x)
     = 3e^2x(A cos x + B sin x) + 4e^2x(– A sin x + B cos x),
4y‘ = 8e^2x(A cos x + B sin x) + 4e^2x(– A sin x + B cos x),
4y = 4e^2x(A cos x + B sin x),
e^2x cos x = y” + 4y‘ + 4y
              = 15e^2x(A cos x + B sin x) + 8e^2x(– A sin x + B cos x)
              = e^2x((15A + 8B) cos x + (– 8A + 15B) sin x),
e^2x((15A + 8B) cos x + (– 8A + 15B) sin x) = e^2x cos x,
(15A + 8B) cos x + (– 8A + 15B) sin x = cos x,

[1] x 8:  120A + 64B = 8,  [3]
[2] x 15:  – 120A + 225B = 0,  [4]
[3] + [4]:  289B = 8, B = 8/289,
[1]:  A = (1 – 8(8/289))/15 = 15/289,
a PS of the NHE is y = e^2x((15/289) cos x + (8/289) sin x).

So the GS of the given NHE is y = c₁e^–2x + c₂xe^–2x + e^2x((15/289) cos x + (8/289) sin x).

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10. Determine the GS of the equation y” + y‘ = 3x².

Solution

a₂ = 3/3 = 1, a₁ = –6(1)/2 = –3, a₀ = –2(–3) = 6,
a PS of the NHE is y = x³ – 3x² + 6x.

So the GS of the given NHE is y = c₁ + c₂e^–x + x³ – 3x² + 6x.

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11. Solve the DE y” + 4y = sin 2x.

Solution

B = 0, A = –1/4,
a PS of the NHE is y = –(1/4)x cos 2x.

Thus the GS of the given NHE is y = c₁ cos 2x + c₂ sin 2x – (1/4)x cos 2x, or y = (– (1/4)x + c₁) cos 2x + c₂ sin 2x.

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12. Find the GS of x” – 4x‘ + 4x = e^2t.

Solution

For the GS of the HE x” – 4x‘ + 4x = 0, we have (–4)² – 4(4) = 0, –(–4)/2 = 2, and the GS of the HE is x = c₁e^2t + c₂te^2t.

For a PS of the given NHE. The functions x = ae^2t and x = ate^2t are PSs of the HE. So let’s try x = at²e^2t. We have:

x = at²e^2t,
x‘ = 2ate^2t +2at²e^2t = (2at² + 2at)e^2t,
x” = (4at + 2a)e^2t + 2(2at² + 2at)e^2t = (4at² + 8at + 2a)e^2t,
–4x‘ = (– 8at² – 8at)e^2t,
4x = 4at²e^2t,
e^2t = x” – 4x‘ + 4x = 2ae^2t,
2ae^2t = e^2t,
2a = 1,
a = 1/2,
a PS of the NHE is x = (1/2)t²e^2t.

Thus the GS of the given NHE is x = c₁e^2t + c₂te^2t + (1/2)t²e^2t, or x = ((1/2)t² + c₂t + c₁)e^2t.

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       y” + by‘ + cy = f (x)

      has a PS that’s a polynomial of degree n.

Solution

Suppose:

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14. Prove that the GS of the NHDE:

Solution

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16.2.1 Equations With Constant Coefficients – Characteristic Equation

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1. Second-Order Linear Homogeneous Differential Equations With     Constant Coefficients

In this section, we’ll use the abbreviations:

“CE”	for	“characteristic equation”,
“DE”	for	“differential equation”,
“GS”	for	“general solution”, and
“PS”	for	“particular solution”.

In the differential equation:

y” + b(x)y‘ + c(x)y = q(x),

For example, the equations:

y” + 3y‘ – 2y = 0     and     y” – 5x²y‘ + 3e^5xy = 0

are linear, while the equations:

y” + 3( y‘)³ – 2y² = 0     and     y”y – 5x²y‘ + 3e^5xy = 0

are non-linear.

The differential equation y” = f (x), as discussed in Section 16.1.1 Part 3, is a second-order linear one with constant
coefficients. For this equation, b(x) = c(x) = 0 and q(x) = f (x).

Definition 1.1 – Second-Order Linear Homogeneous Differential Equations With Constant
                          Coefficients

A differential equation of the form: where b and c are constants, is called a second-order linear homogeneous differential equation with constant coefficients.

When The Coefficient Of y” Isn’t 1

The DE:

ay” + b₁y‘ + c₁y = 0,

where a isn’t 1 or 0, can be transformed into one of the form of Eq. [1.1] by dividing both sides by a:

So y₁ is also a solution of ay” + b₁ y‘ + c₁ y = 0. Thus to solve ay” + b₁ y‘ + c₁ y = 0, where a isn’t 1 or 0, we first divide both
sides of it by a to get the form of Eq. [1.1].

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A Particular Solution

Of course the second-order linear homogeneous DE with constant coefficients:

always has y = 0 as a PS (if y = 0 then y‘ = 0 and y” = 0 and thus y” + by‘ + cy = 0). Of course we’re interested in finding
non-0 solutions of this DE.

By Section 16.1.3 A Particular Case, the GS of the first-order linear DE y‘ + by = 0 is y = Ce^–bx, where C is an arbitrary
constant. So a PS is y = e^–bx. Thus it’s natural to wonder if Eq. [2.1] has a PS of the form:

Characteristic Equations

The quadratic equation:

Definition 2.1 – The Characteristic Equation

Given the differential equation: is called the characteristic equation of the differential equation [2.2].

Solutions Obtained From Two Solutions

So the sum c₁y₁ + c₂y₂ is also a solution. We’ve proved the following theorem.

Theorem 2.1 – Solutions Obtained From Two Solutions

Suppose y1 and y2 are solutions of: Then for any constants c1 and c2, the function: y3 = c1 y1 + c2 y2 is also a solution of Eq. [2.4].

General Solution

Some PSs of the DE y‘ = 2x are y = x², y = x² + 10, or y = x² – 5/4. Its GS is y = x² + C, where C is an arbitrary constant. Recall that the GS of a DE is the solution that represents all the solutions, ie, every PS can be obtained from the GS by giving appropriate value(s) to the constant(s) present in the GS. This definition of course also applies to the GS of a second-order

linear homogeneous DE with constant coefficients.

Linear Independence And General Solution

Let y₁ and y₂ be 2 different solutions of Eq. [2.4]. Then, by Theorem 2.1, for any constants c₁ and c₂ the function c₁y₁ + c₂y₂ is
also a solution. That is, if a function is of the form c₁y₁ + c₂y₂ then it’s a solution. We wonder if the converse is true: if a
function is a solution, is it of the form c₁y₁ + c₂y₂? That is, is every solution of the form c₁y₁ + c₂y₂? That means, is c₁y₁ + c₂y₂
the GS? That is, can every solution be obtained from c₁ y₁ + c₂ y₂ by giving appropriate particular values to c₁ and c₂?

Suppose y₁, y₂, and y₃ are solutions where y₂ is a constant multiple of y₁ but y₃ isn’t a constant multiple of y₁. So y₂ = ky₁,
where k is a constant other than 0 and 1. We have:

c₁y₁ + c₂y₂ = c₁y₁ + c₂(ky₁) = (c₁ + c₂k)y₁ = Ky₁,

where K = c₁ + c₂k, and any solution obtained from y₁ and y₂ can actually be obtained from y₁ alone. Now c₁y₁ + c₃y₃ can’t be
reduced to the form Ly₁, where L is constant, and any solution obtained from y₁ and y₃ must indeed be obtained from both y₁
and y₃; it can’t be obtained just from y₁ alone.

Thus a PS that’s obtained from y₁ and y₃ where y₃ isn’t a constant multiple of y₁ can’t be obtained from y₁ alone, ie it can’t be
obtained from y₁ and y₂ where y₂ is a constant multiple of y₁. Consequently c₁y₁ + c₂y₂ isn’t the GS if y₂ is a constant multiple
of y₁.

Also recall that if two vectors v₁ and v₂ in a plane are linearly independent, then any vector v₃ in the same plane can be
expressed as a linear combination of v₁ and v₂: v₃ = k₁v₁ + k₂v₂ for some constants k₁ and k₂. That is, the linear combination
c₁v₁ + c₂v₂, where c₁ and c₂ are arbitrary constants, represents all the vectors in that plane.

Analogously, two functions f (x) and g(x) are said to be linealy independent if none is a constant multiple of the other, ie if
there doesn’t exist a constant k such that f (x) = kg(x) for all x where both f (x) and g(x) are defined, ie if the ratio f (x)/g(x)
isn’t a constant, or if they have the following property: if c₁ f (x) + c₂ g(x) = 0, where c₁ and c₂ are constants, then c₁ = c₂ = 0.

The answer to our wondering is as follows. If y₁ and y₂ are particular linearly independent solutions of Eq. [2.4], then the GS
of Eq. [2.4] is y = c₁y₁ + c₂y₂, where c₁ and c₂ are arbitrary constants. The GS is a linear combination of the 2 particular linearly independent solutions. The proof of the following theorem is given in a course on differential equations and is omitted

here.

Theorem 2.2 – General Solution

If y1 and y2 are linearly independent particular solutions of the HE: then the general solution of this equation is: y = c1 y1 + c2 y2, where c1 and c2 are arbitrary constants.

Remark 2.1

How about y = c₁ y₁ + c₂ y₂ + C, where C is also an arbitrary constant? Shouldn’t it instead be the general solution? Let’s find
out:

So y = c₁ y₁ + c₂ y₂ + C isn’t the general solution of y” + by‘ + cy = 0. The reason is that the general solution of a
second-order equation involves only two arbitrary constants, not three, and y = c₁ y₁ + c₂ y₂ already involves two, which are c₁
and c₂ .

Remark 2.2

Combining Theorems 2.1 and 2.2 we see that if y₁ and y₂ are solutions of y” + by‘ + cy = 0, then for any constants c₁ and c₂,
c₁ y₁ + c₂ y₂ is also a solution, and additionally if y₁ and y₂ are linearly independent, then for arbitrary constants c₁ and c₂ , c₁ y₁
+ c₂ y₂ is the GS.

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3. The Method Of Characteristic Equation

Recall that to solve a DE is to find its GS. Let’s discuss the solving of our equation:

General Solution If b² – 4c > 0

If b² – 4c > 0, then the CE of Eq. [3.1] has 2 distinct real roots:

where c₁ and c₂ are arbitrary constants.

General Solution If b² – 4c = 0

If b² – 4c = 0, then the CE of Eq. [3.1] has 2 equal real roots:

divide both sides by u‘y₁:

y₂ = uy₁ = (Cx + K)e^–(b/2)x.

Let’s check that y₂ is indeed a solution of Eq. [3.1]:

y = c₁e^–(b/2)x + c₂xe^–(b/2)x,

where c₁ and c₂ are arbitrary constants.

General Solution If b² – 4c < 0

If b² – 4c < 0, then the CE of Eq. [3.1] has 2 imaginary roots:

where c₁ and c₂ are arbitrary constants.

In Section 15.3 Problem & Solution 8, we established Euler formula:

e^ix = cos x + i sin x.

Thus the GS can be written as:

where c₁ and c₂ are arbitray constants.

General Solution Of y” + by‘ + cy = 0

All the above discussion shows that we’ve proved the following theorem. The above procedure relies on the characteristic
equation, and thus is called the method of characteristic equation. Recall that the GS of a DE is such that every PS of the
DE can be obtained from it by assigning appropriate values to the constants present in it.

Theorem 3.1 – General Solution Of  y” + by‘ + cy = 0

Consider the second-order linear homogeneous differential equation with constant coefficients: and let c1 and c2 be arbitrary constants. By the method of characteristic equation:     and the general solution of the differential equation is:

Remarks 3.1

a. Once we’ve found 2 linearly independent solutions of Eq. [3.2], we’ve essentially found all  the solutions of that equation.

b. The GS of Eq. [3.2] always involves the exponential function e^kx, where k is a real number.
c. Eq. [3.2] has infinitely many solutions.

Example 3.1

Find the GS of the DE y” – 3y‘ + 2y = 0.

Solution
(–3)² – 4(2) = 1,

the GS is:

y = c₁e^2x + c₂e^x,

where c₁ and c₂ are arbitrary constants.
EOS

Example 3.2

Note

Recall that to solve a DE is to find its GS.

Solution

where c₁ and c₂ are arbitrary constants.
EOS

Remark 3.2

When the letter x is used for the function, of course we can’t use it also for the variable, we must use another letter. Usually
when the function is x, it’s customary that the variable is t.

Example 3.3

Solve the equation w” + w‘ + w = 0.

Solution
1² – 4(1) = –3,

where c₁ and c₂ are arbitrary constants.
EOS

Remark 3.3

When a letter other than x, whatever it is, is used for the function, we can always use the letter x for the variable.

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4. For Initial-Value Problems

As seen in Section 16.1.1 Remarks 4.1, an initial-value problem involving a first-order DE specifies 1 initial condition. As seen in
the same section Remarks 4.2, an initial-value problem involving a second-order DE specifies 2 initial conditions. The DE y” +
by‘ + cy is of second order, and so an intial-value problem involving it specifies 2 initial conditions. This fact is also clear from
the form of the equation’s GS: y = c₁y₁ + c₂y₂, where there are 2 constants c₁ and c₂ to be determined to get a PS. To solve an
initial-value problem is to find the PS of the DE that satisfies the conditions.

Example 4.1

Solve the initial-value problem:

Solution
3² – 4(–10) = 49,

EOS

Remark 4.1

In Example 3.1, we differentiate the GS, use the condition y‘(0) = 3 to get an equation in c₁ and c₂, use the condition y(0) = 1
to get another equation in c₁ and c₂, obtain a system of 2 simultaneous equations in 2 unknowns c₁ and c₂, solve it, obtain a
unique set of solutions, and thus obtain a unique PS of the given initial-value problem.

In general, the second-order linear DE y” + by‘ + cy = 0 has infinitely many PSs, but when 2 initial conditions are imposed, there’s only 1 PS that satisfies the conditions. Note that for a second-order equation, there are 2 initial conditions, which are on

y‘ and y (= y⁽⁰⁾).

Theorem 4.1 – Existence And Uniqueness Of A Solution

The solution of the initial-value problem: where x0, k0, and k1 are real numbers, exists and is unique.

This theorem is a special case where n = 2 of the general theorem that asserts that the solution of an initial-value problem
involving an nth-order linear DE:

y⁽ⁿ⁾ + p_n–1(x)y^(n–1) + p_n–2(x)y^(n–2) + … + p₂(x)y” + p₁(x)y‘ + p₀(x)y = q(x)

and n initial conditions:

y(x₀) = k₀,     y‘(x₀) = k₁,     y”(x₀) = k₂,     …,     y^(n–2)(x₀) = k_n–2,     y^(n–1)(x₀) = k_n–1,

exists and is unique provided the functions p_i(x)’s and q(x) are continuous on a neighborhood of x₀. This general theorem is
presented and proved in a course on differential equations. Remark that for an nth-order equation, there are n conditions,
which are on y (= y⁽⁰⁾) thru to y^(n–1).

Example 4.2

Solve the following initial-value problem:

Solution
0² – 4(1) = – 4,

c₁ = –1,

the desired PS is y = (–1) cos x + (0) sin x, or y = – cos x.
EOS

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1. Find the GS of the DE 4y” + 12y‘ + 5y = 0.

Solution

y = c₁e^–(1/2)x + c₂e^–(5/2)x.

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2. Solve the DE x” – 8x‘ + 16x = 0.

Solution

(–8)² – 4(16) = 0,

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3. Solve the equation y” + 3y = 0.

Solution

0² – 4(3) = – 12,

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4. Find the solution of the initial-value problem:

Solution

(–5)² – 4(6) = 1,

y = c₁e^3x + c₂e^2x,
1 = y(0) = c₁e⁰ + c₂e⁰ = c₁ + c₂,     c₁ + c₂ = 1,  [1]
y‘ = 3c₁e^3x + 2c₂e^2x,
–1 = y‘(0) = 3c₁e⁰ + 2c₂e⁰ = 3c₁ + 2c₂,     3c₁ + 2c₂ = –1,  [2]
[1] x (–3):  – 3c₁ – 3c₂ = –3,  [3]
[2] + [3]:  – c₂ = – 4,
c₂ = 4,
[1]:  c₁ = 1 – c₂ = 1 – 4 = –3,
y = – 3e^3x + 4e^2x, or y = 4e^2x – 3e^3x.

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5. Solve the following initial-value problem:

Solution

4v” + 20v‘ + 25v = 0,

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6. Solve this initial-value problem:

Solution

0² – 4(4) = –16,

c₂ = –4/2 = –2,

y = 3 cos 2x – 2 sin 2x.

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7. Consider the DE:

    y‘ + y² + cy + d = 0,

    where c and d are constants. It’s a first-order non-linear equation. Observe that the exponents of y decrease from 2 to 1 to
    0 and that the coefficient of y² is 1.

a. Transform the given y-equation into a second-order linear homogeneous z-equation with constant coefficients by letting z
    be a function such that z‘/z = y.

b. Prove that if z is the GS of the z-equation found in part a and y = z‘/z, then y is the GS of the y-equation.

c. Prove that the GS of the y-equation, which is of first-order, indeed involves only 1 arbitrary constant.

Solution

a. Let z be a function such that z‘/z = y. Then:

    Thus, by part a, w is a solution of the z-equation, and, by part b, w is a PS of the z-equation, for if it’s the GS of the
    z-equation then v would be the GS of the y-equation, a contradiction. Since z is the GS of the z-equation which is
    second-order linear homogeneous with constant coefficients, we have:

    z = c₁z₁ + c₂z₂,

    where z₁ and z₂ are linearly independent PSs of the z-equation and c₁ and c₂ are arbitrary constants. As a consequence:

    w = k₁z₁ + k₂z₂,

    where k₁ and k₂ are particular constants. Then:

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16.2.2 Equations With Variable Coefficients – Reduction Of Order

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1. Second-Order Linear Homogeneous Differential Equations With     Variable Coefficients

In this section, we’ll use the abbreviations:

“DE”	for	“differential equation”,
“GS”	for	“general solution”,
“HDE”	for	“homogeneous differential equation”, and
“HE”	for	“homogeneous equation”.

Recall that a second-order linear homogeneous differential equation with constant coefficients is one of the form:

y” + by‘ + cy = 0,

where the coefficients b and c are constants. When b is replaced by a non-constant function b(x) or c is replaced by a
non-constant function c(x) or both, one or both coefficients of course vary, they become variable. So naturally an equation of
the form:

y” + b(x) y‘ + c(x) y = 0,

where b(x) or c(x) or both are non-constant functions of x, is said to be an equation with variable coefficients. Note that x is
the independent variable of the function y, and thus of the functions y‘ and y” also. All the functions in the equation are of the same variable.

Definition 1.1 – Second-Order Linear Homogeneous Differential Equations With Variable
                          Coefficients

A differential equation of the form: where b(x) or c(x) or both are non-constant functions, is called a second-order linear homogeneous differential equation with variable coefficients.

Recall that while the equation is linear, each function y, y‘, and y” doesn’t have to be linear. For example, y = 2x + 3 is linear
while y = x² + 3 and y = e^2x + 3 aren’t.

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Just like the GS of the constant-coefficients HE:

y” + by‘ + cy = 0

as presented in Section 16.2.1 Theorem 2.2, the GS of the variable-coefficients HE:

y” + b(x) y‘ + c(x) y = 0

is the linear combination of any two of its linearly independent PSs. The proof of the following theorem is given in a course on
differential equations and is omitted here.

Theorem 2.1 – General Solution

If y1 and y2 are linearly independent particular solutions of the HE: y” + b(x) y‘ + c(x) y = 0, then the GS of this equation is: y = c1 y1 + c2 y2, where c1 and c2 are arbitrary constants.

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3. The Method Of Reduction Of Order

Recall that to solve a DE is to find its GS. Let’s discuss the solving of our equation:

By Theorem 2.1, we obtain the GS of Eq. [3.1] provided we get two linearly independent solutions y₁ and y₂ of this equation.
The GS is then given by the linear combination:

y = c₁ y₁ + c₂ y₂,

where c₁ and c₂ are arbitrary constants. Clearly y₁ = 0 is a solution. But it’s not linearly independent of any other solution (0 =
0y₂ whatever y₂ is). So of course we’re interested in non-0 solutions.

Unlike the case of constant-coefficients HEs as developed in Section 16.2.1 Part 3, there’s no standard procedure to find two
linearly independent solutions y₁ and y₂ of the variable-coefficients HE [3.1]. However, there’s a general procedure for finding
another solution if one solution is known. It’s often possible to find one solution by inspecting the equation or by trial and error.

Let y₁ be a non-0 solution of Eq. [3.1]. If y₂ is another solution that’s linearly independent of y₁, then y₂/y₁ isn’t a constant.
Thus:

Indeed y₂ is a solution. Now let’s verify that it’s linearly independent of y₁. We have:

We’ve reduced a second-order equation to a first-order equation. The order of the derivative is reduced. So this approach of
using one solution to find another is called the method of reduction of order.

We’ve of course proved the following theorem.

Theorem 3.1 –A Solution By The Method Of Reduction Of Order

Suppose y1 is a solution of the DE: where c1 and c2 are arbitrary constants.

Remark 3.1

Example 3.1

Consider the DE:

x²y” – xy‘ + y = 0,     x > 0,

a. Verify that y₁ = x is a solution.

b. Find a second solution y₂ that’s linearly independent of y₁ by using the formula obtained by the method of reduction of order.
c. Find y₂ by imitating the method of reduction of order itself, not using the formula obtained by it.
d. Are the answers in parts b and c the same?
e. Verify that y₂ is indeed a solution that’s linearly independent of y₁.

Solution

    xu” + u‘ = 0,
    divide both sides by xu‘:

     y₂ = uy₁ = (ln x) x = x ln x.

d. Yes.

    Since ln x isn’t a constant, y₂ is linearly independent of y₁. Thus indeed y₂ is a solution that’s linearly independent of y₁.

EOS

Remarks 3.2 – About Example 3.1

a. In part b, we utilize the formula for y₂, which contains the function b(x), so we should write the given equation in the form
    y” + b(x) y‘ + c(x) y = 0 (coefficient of y” is 1), so that it’s clear what b(x) is. Caution: b(x) isn’t –x, it’s –1/x.

b. Part c illustrates the following general procedure of the method of reduction of order:

    – let y₂ = uy₁ = g(x)u, where y₁ = g(x), use g(x) only, not y₁,
    – substitute y₂ = g(x)u into the original equation,
    – transform the term containing u” in an equation into the term u”/u‘ by division,

    – let v = u‘,
    – find v, then u, then y₂, see Fig. 3.1.

Process From y2 To u To v Then Back To u To y2.

d. If we’re not asked to verify a solution, then we don’t have to verify.

Remark 3.3

It’s recommended that you remember the procedure of the method of reduction of order, as outlined in part b of Remarks 3.2 above. In case you forget its formula or don’t remember the formula clearly, you can utilize the procedure of the method.

Sometimes you may even be required to use the method itself, not the formula obtained by it.

Choosing y₂

Suppose it’s determined that y₂ is a constant multiple of a function, ie that it’s of the form y₂ = k f (x), where k is a non-0
constant. For example, y₂ = 5x³ or y₂ = –3(sin 2x + 7e^4x). We now verify that f (x) is also a solution that’s linearly independent
of y₁. We have:

which isn’t constant, since y₂/y₁ isn’t. Thus f (x) is linearly independent of y₁. That completes the verification.

We’ve shown that if k f (x), where k is a non-0 constant, is a solution that’s linearly independent of y₁, then so is f (x). We
conclude that if we determine that y₂ = k f (x), then it’s possible to simply choose y₂ = f (x).

Let’s look at this possibility from another angle. If y₂ = k f (x), then the GS is y = c₁y₁ + c₂k f (x). Re-using the letter c₂ for the
product c₂k, which is valid because this product, just like the factor c₂ in it, is an arbitrary constant, the GS is y = c₁y₁ +
c₂ f (x). This is the same as simply choosing y₂ = f (x).

Example 3.2

One solution of the DE:

xy” – y‘ + 4x³y = 0,     x > 0,

is y₁ = sin (x²). Solve the equation.

Solution
xy” – y‘ + 4x³y = 0,

Let u = x². So du = 2x dx. Then:

So, choosing y₂ = cos (x²), the GS is y = c₁ sin (x²) + c₂ cos (x²).
EOS

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4. For Constant-Coefficients Equations

The procedure of the method of reduction of order doesn’t require that the coefficients be variable functions. So the method
also applies to constant-coefficients equations.

Example 4.1

Consider the constant-coefficients DE:

y” + 5y‘ + 6y = 0.

a. Solve it by using the method of characteristic equation.
b. Given that one solution is y₁ = e^–2x, solve it by using the method of reduction of order.
c. Are the answers in parts a and b the same?

Solution

     So, choosing y₂ = e^–3x, the GS is y = c₁e^–2x + c₂e^–3x.

c. Yes.
EOS

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1. Consider the Legendre equation of order 1:

    (1 – x²) y” – 2xy‘ + 2y = 0,     |x| < 1.

    a. Verify that y₁ = x is a solution.

    b. Find a second solution y₂ that’s linearly independent of y₁ by using the formula obtained by the method of reduction of order.
    c. Find y₂ by imitating the method of reduction of order itself, not using the formula obtained by it.
    d. Are the answers in parts b and c the same?
    e. Verify that y₂ is indeed a solution that’s linearly independent of y₁.

Solution

    Let u = 1 – x². So du = –2x dx. Then:

    Using the method of partial fractions we have:

     C = 0,
     D = –1,
     0 = A + B + C = A + B + 0 = A + B,     B = –A,
     0 = – A + B + D = – A – A – 1 = – 2A – 1,     A = –1/2,
     B = –(–1/2) = 1/2,

    Using the method of partial fractions we have:

    Using the method of partial fractions we have:

d. Yes.

e) Let’s show that y₂ is a solution:

    So indeed y₂ is a solution. Now let’s show that it’s linearly independent of y₁:

    which clearly isn’t constant. Thus indeed y₂ is linearly independent of y₁.

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2. Consider the DE:

    x²y” – 2xy‘ + (x² + 2) y = 0,     x > 0.

    a. Verify that y₁ = x sin x is a solution.
    b. Solve the equation.

Solution

    Choose y₂ = x cos x. So the GS of the given equation is y = c₁ x sin x + c₂ x cos x, or:

    y = x(A cos x + B sin x),

    where A = c₂ and B = c₁ are arbitrary constants.

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3. Consider the constant-coefficients DE:

    y” + y‘ + 7y = 0.

a. Verify that:

    is a solution.

b. Use y₁ to solve the equation.

Note

We’re required to use y₁ to solve the equation. So we can’t apply the method of characteristic equation, which doesn’t use y₁.
We must apply the method of reduction of order, which uses y₁ to find y₂.

Solution

    Indeed y₁ is a solution.

b. Let y₂ be another solution that’s linearly independent of y₁. Then by the method of reduction of order we have:

    where A = c₂ and B = c₁ are arbitrary constants.

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4. Consider the Bessel DE:

    x²y” + xy‘ + (x² – p²) y = 0,     p a constant.

    For p = 1/2 and x > 0:

b. Solve the equation.

Solution

a. When p = 1/2 the given equation becomes:

    where A and B are arbitrary constants.

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5. Given the DE:

    xy” – (x + n) y‘ + ny = 0,     x > 0,     integer constant n > 0,

    verify that y₁ = e^x is a solution and determine another solution linearly independent of y₁.

Solution

Let’s verify that y₁ is a solution. We have:

Thus:

I_n = – xⁿe^–x + nI_n–1

   = – e^–x(xⁿ) + n(– x^n–1e^–x + (n – 1)I_n–2)

   = – e^–x(xⁿ + nx^n–1) + n(n – 1)(– x^n–2e^–x + (n – 2)I_n–3)

   = – e^–x(xⁿ + nx^n–1 + n(n – 1)x^n–2) + n(n – 1)(n – 2)(– x^n–3e^–x + (n – 3)I_n–4)

   = – e^–x(xⁿ + nx^n–1 + n(n – 1)x^n–2 + n(n – 1)(n – 2)x^n–3) + n(n – 1)(n –2)(n – 3)(– x^n–4e^–x + (n – 4)I_n–5)

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